Estimate Instantaneous Velocity

In summary: You have the equation: y(t)=48t−16t2When I plug 48(t)-16(t)^2 into \dfrac{y(2+h)-y(2)}{(2+h)-2}, I don't understand how to do it. Am I actually plugging in 48(t)-16(t)^2 into the y values of the other equation? Something like...y(t)=48t−16t2...or is there a different way to do it?
  • #1
FritoTaco
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Homework Statement


If a ball is thrown into the air with a velocity of 48 ft/s, its height in feet t seconds later is given by y = 48t − 16t2.
(a) Find the average velocity for the time period beginning when t = 2 and lasting for each of the following.

(i) 0.5 seconds

(ii) 0.1 seconds

(iii) 0.05 seconds

(iv) 0.01 seconds

(b) Estimate the instantaneous velocity when t = 2.

Homework Equations



[itex]Average Velocity =\dfrac{\triangle y}{\triangle t}[/itex] (Change in position over change in time)

[itex]m=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}[/itex]

The Attempt at a Solution



Set: [itex]t = 2[/itex] and [itex]t = 2 + h[/itex] (let h be the time difference) [itex]\ne 0[/itex]

[itex]t = 2[/itex] is [itex]y_{1},x_{1}[/itex] and [itex]t = 2 + h[/itex] is [itex]y_{2},x_{2}[/itex]

Average Velocity = [itex]\dfrac{y(2+h)-y(2)}{(2+h)-2}[/itex]

Now with [itex]y=48t-16t^2[/itex], we will replace [itex]t[/itex] with [itex]2[/itex] and [itex]2 + h[/itex].

[itex]\dfrac{48(2+h)-16(2+h)^2-[48(2)-16(2)^2)]}{h}[/itex]

[itex]\dfrac{96+48h-16(4+4h+h^2)-96+64}{h}[/itex]

[itex]\dfrac{48h-64-64h-16h^2+64}{h}[/itex]

[itex]\dfrac{-16h^2-16h}{h}[/itex]

[itex]\dfrac{h(-16h-16)}{h}[/itex]

Average Velocity =[itex]-16h-16[/itex]

Then I plug in (i) 0.5 into the equation like this: [itex]-16(0.5)-16=24[/itex]
I do this for the rest of them and get correct answers according to the online "WebAssign" website. Now I'm stuck on (b) "Estimate the instantaneous velocity when t = 2." You can look at my picture, I tried plugging in values near 2 like 2.1 and 2.01, etc, into [itex]y=48t-16t^2[/itex] What am I doing wrong? So as I get closer to 2 I get closer to the slope of 31.999 which then you make an educated guess (by rounding) and get 32. But this is wrong. I don't know why? Also, I heard there is another method for finding average velocity, my book doesn't have it--anyone know it? Much appreciation.
 

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  • #2
FritoTaco said:

Homework Statement


If a ball is thrown into the air with a velocity of 48 ft/s, its height in feet t seconds later is given by y = 48t − 16t2.
(a) Find the average velocity for the time period beginning when t = 2 and lasting for each of the following.

(i) 0.5 seconds

(ii) 0.1 seconds

(iii) 0.05 seconds

(iv) 0.01 seconds

(b) Estimate the instantaneous velocity when t = 2.

Homework Equations



[itex]Average Velocity =\dfrac{\triangle y}{\triangle t}[/itex] (Change in position over change in time)

[itex]m=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}[/itex]

The Attempt at a Solution



Set: [itex]t = 2[/itex] and [itex]t = 2 + h[/itex] (let h be the time difference) [itex]\ne 0[/itex]

[itex]t = 2[/itex] is [itex]y_{1},x_{1}[/itex] and [itex]t = 2 + h[/itex] is [itex]y_{2},x_{2}[/itex]

Average Velocity = [itex]\dfrac{y(2+h)-y(2)}{(2+h)-2}[/itex]

Now with [itex]y=48t-16t^2[/itex], we will replace [itex]t[/itex] with [itex]2[/itex] and [itex]2 + h[/itex].

[itex]\dfrac{48(2+h)-16(2+h)^2-[48(2)-16(2)^2)]}{h}[/itex]

[itex]\dfrac{96+48h-16(4+4h+h^2)-96+64}{h}[/itex]

[itex]\dfrac{48h-64-64h-16h^2+64}{h}[/itex]

[itex]\dfrac{-16h^2-16h}{h}[/itex]

[itex]\dfrac{h(-16h-16)}{h}[/itex]

Average Velocity =[itex]-16h-16[/itex]

Then I plug in (i) 0.5 into the equation like this: [itex]-16(0.5)-16=24[/itex]
I do this for the rest of them and get correct answers according to the online "WebAssign" website. Now I'm stuck on (b) "Estimate the instantaneous velocity when t = 2." You can look at my picture, I tried plugging in values near 2 like 2.1 and 2.01, etc, into [itex]y=48t-16t^2[/itex] What am I doing wrong? So as I get closer to 2 I get closer to the slope of 31.999 which then you make an educated guess (by rounding) and get 32. But this is wrong. I don't know why? Also, I heard there is another method for finding average velocity, my book doesn't have it--anyone know it? Much appreciation.
For the instantaneous velocity near t = 2, look at the change in position divided by the change in time.

Look at ##\frac{y(2 + h) - y(2)}{2 + h - 2}##, for some values of h that are close to 0, say h = .1, h = .01, h = .001.
You should get a value close to -16 ft/sec.
 
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  • #3
My teacher also said use values like 0.1, 0.01, etc. Why? I don't get why we're using those values? Also, how do we make the equation where we can plug in those values? from using [itex]\dfrac{y(2+h)-y(2)}{(2+h)-(2)}[/itex]
 
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  • #4
FritoTaco said:
My teacher also said use values like 0.1, 0.01, etc. Why? I don't get why we're using those values?
Because you want the time interval to be very short so as to get a closer estimate of the instantaneous velocity.
FritoTaco said:
Also, how do we make the equation where we can plug in those values? from using [itex]\dfrac{y(2+h)-y(2)}{(2+h)-(2)}[/itex]
You have the equation: ##y(t) = 48t - 16t^2##
 
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  • #5
Mark44 said:
Because you want the time interval to be very short so as to get a closer estimate of the instantaneous velocity.

This is where I don't understand. It says find instantaneous velocity when t = 2. Okay, so don't we want to use values close to 2? Like 2.1, 2.01. So if it was like, "find instantaneous velocity closer to when t = 5", would I still use 0.1, 0.01, etc? Because that means we are 0.1 from 5 or 0.01 from 5?

Mark44 said:
You have the equation: y(t)=48t−16t2

When I plug [itex]48(t)-16(t)^2[/itex] into [itex]\dfrac{y(2+h)-y(2)}{(2+h)-2}[/itex], I don't understand how to do it. Am I actually plugging in [itex]48(t)-16(t)^2[/itex] into the y values of the other equation? Something like this:

[itex]\dfrac{48t-16t^2(2+h)-48t-16t^2(2)}{2+h-2}[/itex]? Then distribute and simplify it down more?
 
  • #6
FritoTaco said:
This is where I don't understand. It says find instantaneous velocity when t = 2. Okay, so don't we want to use values close to 2? Like 2.1, 2.01. So if it was like, "find instantaneous velocity closer to when t = 5", would I still use 0.1, 0.01, etc? Because that means we are 0.1 from 5 or 0.01 from 5?
Yes. If h = .01, then 2 + h = 2 + .01 = 2.01
FritoTaco said:
When I plug [itex]48(t)-16(t)^2[/itex] into [itex]\dfrac{y(2+h)-y(2)}{(2+h)-2}[/itex], I don't understand how to do it. Am I actually plugging in [itex]48(t)-16(t)^2[/itex] into the y values of the other equation? Something like this:
[itex]\dfrac{48t-16t^2(2+h)-48t-16t^2(2)}{2+h-2}[/itex]? Then distribute and simplify it down more?
No, you are "plugging in" 2 + h and 2, respectively, into your height function y(t) = 48t - 16t2
For y(2 + h), replace t in your formula with 2 + h.
For y(2), replace t with 2.
Then simplify the result.
 
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  • #7
Mark44 said:
No, you are "plugging in" 2 + h and 2, respectively, into your height function y(t) = 48t - 16t2
For y(2 + h), replace t in your formula with 2 + h.
For y(2), replace t with 2.
Then simplify the result.

[itex]48(2+h)-16(2)^2[/itex]

[itex]96+48h-64[/itex]

[itex]48h-32[/itex]

Am I doing this right?
 
  • #8
FritoTaco said:
[itex]48(2+h)-16(2)^2[/itex]
I don't know what you're doing above.
FritoTaco said:
[itex]96+48h-64[/itex]

[itex]48h-32[/itex]

Am I doing this right?
No. This is what you should be evaluating: ##\frac{y(2+h)-y(2)}{(2+h)-2}##, using ##y(t) = 48t - 16t^2##. The quotient here gives you the average speed between time t = 2 + h and time t = 2.
Edit: Fixed a typo I had in the y(t) formula.
 
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  • #9
Mark44 said:
No. This is what you should be evaluating: y(2+h)−y(2)(2+h)−2y(2+h)−y(2)(2+h)−2\frac{y(2+h)-y(2)}{(2+h)-2}, using y(t)=48t−32t2y(t)=48t−32t2y(t) = 48t - 32t^2. The quotient here gives you the average speed between time t = 2 + h and time t = 2.

How did you get [itex]y(t)=48t-32t^2[/itex]?

Also, in above when you said "For y(2 + h), replace t in your formula with 2 + h. For y(2), replace t with 2." Isn't that what I did and got the [itex]48h - 32[/itex]?
 
  • #10
FritoTaco said:
How did you get [itex]y(t)=48t-32t^2[/itex]?
I miswrote 32t2 instead of 16t2. I have fixed my typo.

FritoTaco said:
Also, in above when you said "For y(2 + h), replace t in your formula with 2 + h. For y(2), replace t with 2." Isn't that what I did and got the [itex]48h - 32[/itex]?
Whatever you did isn't right. y(2 + h) gives you two terms, and y(2) also gives you two terms. Also, you have 2 + h - 2 in the denominator.
 
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  • #11
FritoTaco said:
Also, in above when you said "For y(2 + h), replace t in your formula with 2 + h. For y(2), replace t with 2." Isn't that what I did and got the [itex]48h - 32[/itex]?
Let's back up a few steps. You need to evaluate ##\frac{y(2 + h) - y(2)}{2 + h - 2}##, using ##y(t) = 48t - 16t^2##.

What do you get for y(2 + h)?
What do you get for y(2)?
 
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  • #12
Am I placing [itex]48t-16t^2[/itex] into the y part of both [itex]y(2+h)[/itex] and [itex]y(2)[/itex]?

Like this maybe:

[itex]\dfrac{48(2+h)-16(2+h)^2-[48(2)-16(2)^2]}{(2+h)-2}[/itex]

[itex]\dfrac{96+48h-16(h^2+4h+4)-96+64}{(2+h)-2}[/itex]

[itex]\dfrac{96+48h-16h^2-64h-64-96+64}{(2+h)-2}[/itex]

[itex]\dfrac{-16h^2-16h}{(2+h)-2}[/itex]

I'm pretty sure something isn't right but I gave it a go here.
 
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  • #13
FritoTaco said:
Am I placing [itex]48t-16t^2[/itex] into the y part of both [itex]y(2+h)[/itex] and [itex]y(2)[/itex]?

Like this maybe:

[itex]\dfrac{48(2+h)-16(2+h)^2-[48(2)-16(2)^2]}{(2+h)-2}[/itex]

[itex]\dfrac{96+48h-16(h^2+4h+4)-96+64}{(2+h)-2}[/itex]

[itex]\dfrac{96+48h-16h^2-64h-64-96+64}{(2+h)-2}[/itex]

[itex]\dfrac{-16h^2-16h}{(2+h)-2}[/itex]

I'm pretty sure something isn't right but I gave it a go here.
This is correct, but you should simplify the denominator.

When you've done that, estimate the velocity at a few times close to 2 sec., such as 2.1 (h = .1), 2.01 (h = .01), and however more you need to do.
 
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  • #14
Mark44 said:
This is correct, but you should simplify the denominator.

When you've done that, estimate the velocity at a few times close to 2 sec., such as 2.1 (h = .1), 2.01 (h = .01), and however more you need to do.

Will it just be [itex]\dfrac{-16h^2-16h}{h}[/itex]?
 
  • #15
FritoTaco said:
Will it just be [itex]\dfrac{-16h^2-16h}{h}[/itex]?
And simplfy that...
 
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  • #16
Like this:

[itex]\dfrac{-16h(h+1)}{h}[/itex] Factor

[itex]16(h+1)[/itex] Cancel out the h
 
  • #17
FritoTaco said:
Like this:

[itex]\dfrac{-16h(h+1)}{h}[/itex] Factor

[itex]16(h+1)[/itex] Cancel out the h
Where did the minus sign go?

When you get that straigtened out, pick a few (small) values of h, as described in parts i) through iv) of the first post, and then estimate the velocity at t = 2 seconds.
 
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  • #18
Oh, I forgot about that. It should've been there as [itex]-16(h+1)[/itex]

i.)0.5

[itex]-16(0.5+1) = -24 ft/s[/itex]​

ii.)0.1

[itex]-16(0.1+1) = -17.6 ft/s[/itex]​

iii.)0.05

[itex]-16(0.05+1) = -16.8 ft/s[/itex]​

iv.)0.01

[itex]-16(0.01+1) = 1-6.16 ft/s[/itex]​

b.) Estimate the instantaneous velocity when t = 2.

[itex]-16(0.1+1) = -17.6 ft/s[/itex]

[itex]-16(0.01+1) = -16.16 ft/s[/itex]

[itex]-16(0.001+1) = -16.08 ft/s[/itex]

Estimate: [itex]-16 ft/s[/itex].

I entered that in and got it right! Thank you very much for your help!
 
  • #19
FritoTaco said:
iv.)0.01

[itex]-16(0.01+1) = 1-6.16 ft/s[/itex]
Surely, you mean -16.16 ft/sec
FritoTaco said:
b.) Estimate the instantaneous velocity when t = 2.

[itex]-16(0.1+1) = -17.6 ft/s[/itex]

[itex]-16(0.01+1) = -16.16 ft/s[/itex]

[itex]-16(0.001+1) = -16.08 ft/s[/itex]

Estimate: [itex]-16 ft/s[/itex].

I entered that in and got it right! Thank you very much for your help!
 
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  • #20
Oh, oops. Yeah, -16.16ft/s.
 

FAQ: Estimate Instantaneous Velocity

1. What is instantaneous velocity?

Instantaneous velocity is the velocity of an object at a specific moment in time. It is the rate at which an object is changing its position at that moment, and is typically measured in meters per second (m/s).

2. How is instantaneous velocity different from average velocity?

Instantaneous velocity is calculated at a specific moment in time, while average velocity is calculated over a period of time. Average velocity takes into account the total distance traveled and the total time taken, while instantaneous velocity only considers the velocity at that particular moment.

3. How is instantaneous velocity calculated?

Instantaneous velocity can be calculated by taking the derivative of the position function with respect to time. This can be represented as the limit of the average velocity as the time interval approaches zero.

4. Can instantaneous velocity be negative?

Yes, instantaneous velocity can be negative. This indicates that the object is moving in the opposite direction of its positive velocity. For example, if a car is traveling forward with a velocity of 20 m/s and then suddenly stops, its instantaneous velocity at that moment would be -20 m/s.

5. How is instantaneous velocity used in real-world applications?

Instantaneous velocity is used in various fields, such as physics, engineering, and sports. In physics, it is used to analyze the motion of objects and understand their acceleration. In engineering, it is used to design machines and structures that require precise motion control. In sports, it is used to track the performance of athletes and analyze their movements.

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